THE TRISECTION OF ANY ANGLE This article has been written during Markos Georgallides : Tel-00357 -99 634628 last month of 2010 and is solely for Civil Engineer(NATUA) : Fax-00357-24 653551 my own amusement and private study 38 , Z .Kitieos St , 6022 , Larnaca and also to all those readers that believe Expelled from Famagusta town occupied to Euclid Geometry as the model of by the Barbaric Turks . all nature . A meter also for testing sufficiency of Geometries . Email < > 1. Archimedes method 2. Pappus method Consider the angle < AOB . Draw circle ( O , OA ) with its center at the vertex O and produce side BO to D . Insert a straight line AD so point C is on the circle and point D on line BO and length DC such that it is equal to the radius of the circle . 1
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THE TRISECTION OF ANY ANGLE

This article has been written during Markos Georgallides : Tel-00357 -99 634628last month of 2010 and is solely for Civil Engineer(NATUA) : Fax-00357-24 653551 my own amusement and private study 38 , Z .Kitieos St , 6022 , Larnaca and also to all those readers that believe Expelled from Famagusta town occupied to Euclid Geometry as the model of by the Barbaric Turks . all nature . A meter also for testing sufficiency of Geometries . Email < [email protected] >

1. Archimedes method 2. Pappus method

Consider the angle < AOB .

Draw circle ( O , OA ) with its center at the vertex O and produce side BO to D . Insert a straight line AD so point C is on the circle and point D on line BO and length DC such that it is equal to the radius of the circle .

Proof : Since CD = CO then triangle CDO is isosceles and angle < CDO = COD The external angle OCA of triangle CDO is < OCA = CDO + COD = 2. CDO and equal to angle ADO and since angle < OAC = OCA then < OAC = 2.ODA The external angle AOB of triangle OAD is < AOB = OAD + ODA = 2. ODA + ODA = 3 . ODA

2. Pappus method :

It is a slightly different of Archimedes method can be reduced to a neusis as follows Consider the angle < AOB . Draw AB perpendicular to OB .Complete rectangle ABOC . Produce the side CA to E .Insert a straight line ED of given length 2 .OA between AE and AB in such a way that ED verges towards O . Then angle < AOB = 3 . DOB

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3. The Present method :

We extend Archimedes method as follows :

a . ( F. 4 ) Given an angle < AOB = AOC = 90▫ 1. Draw circle ( A , AO = OA ) with its center at the vertex A intersecting circle ( O , OA = AO ) at the points A1 , A2 respectively . 2 . Produce line AA1 at C so that A1C = A1 A = AO and draw AD // OB . 3 . Draw CD perpendicular to AD and complete rectangle AOCD . 4 . Point F΄ is such that OF΄ = 2 . OA

b . ( F. 5-6-7 ) Given an angle < AOB < 90▫ 1. Draw AD parallel to OB .2. Draw circle ( A , AO = OA ) with its center at the vertex A intersecting circle ( O , OA = AO ) at the points A1 , A2 .3. Produce line A A1 at D1 so that A1D1 = A1 A = OA . 4 . Point F is such that OF = 2 . OA = 2 . OAo . 5 . Draw CD perpendicular to AD and complete rectangle A΄OCD .6 . Draw Ao E Parallel to A΄ C at point E ( or sliding E on OC ) .7 . Draw AoE΄ parallel to OB and complete rectangle AoOEE΄ .8 . Draw AF intersecting circle ( O , OA ) at point F1 and insert on AF segment F1 F2 equal to OA → F1 F2 = OA . 9 . Draw AE intersecting circle ( O , OA ) at point E1 and insert on AE segment E1 E2 equal to OA → E1 E2 = OA = F1 F2 .

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Show that :

a) For all angles equal to 90▫ Points C and E are at a constant distance OC = OA . √ 3 and OE = OAo . √ 3 , from vertices O , and also A΄C // AoE . b) The geometrical locus of points C , E is the perpendicular CD , EE΄ . c) All equal circles with their center at the vertices O , A and radius OA = AO have the same geometrical locus EE΄ ┴ OE for all points A on AD , or All radius of equal circles drawn at the points of intersection with its Centers at the vertices O , A and radius OA = AO lie on CD , EE΄ . d) Angle < D1OA is always equal to 90▫ and angle AOB is created by rotation of the right-angled triangle AOD1 through vertex O . e ) Angle < AOB is created in two ways , By constructing circle ( O , OA = OAo ) and by sliding of point A΄ on line A΄ D Parallel to OB from point A΄ to A . f) The rotation of lines AE , AF on circle ( O , OA = OAo ) from point E to point F which lines intersect circle ( O , OA ) at the points E1 , F1 respectively , fixes a point G on line EF and a point G1 common to line AG and to the circle ( O , OA ) such that G G1 = OA .

Proof :

a ) .. ( F3 , F4 )

Let OA be one-dimentional Unit perpendicular to OB such that angle < AOB = AOC = 90▫Draw the equal circles ( O,OA ) , ( A , AO ) and let points A1 , A2 be the points of intersection . Produce AA1 to C . Since triangle AOA1 has all sides equal to OA ( AA1 = AO = OA1 ) then it is an equilateral triangle and angle < A1AO = 60 ▫Since Angle < CAO = 60 ▫ and AC = 2. OA then triangle ACO is right-angled and angle < AOC = 90▫ , so the angle ACO = 30 ▫Complete rectangle AOCD

Angle < ADO = 180 – 90 – 60 = 30 ▫ = ACO = 90 ▫ / 3 = 30 ▫From Pythagoras theorem AC² = AO² + OC² or OC ² = 4.OA² - OA² = 3. OA²

and OC = OA . √ 3 . For OA = OAo then AoE = 2. OAo and OE = OAo . √ 3 . Since OC / OE = OA / OAo → then line CA ΄ is parallel to EAo

b ) .. ( F5 , F6 )

Triangle OAA1 is isosceles , therefore angle < A1AO = 60 ▫ . Since A1A = A1D 1 and angle < A1AO = 60 ▫ then triangle AOD 1 is right-angle triangle and angles < D 1OA = 90 ▫ , angle < OD 1A = 30 ▫ . Since the circle of diameter D1A passes through point O and also through the foot of the perpendicular from point D1 to AD , and since also ODA = ODA΄ = 30 ▫

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then this foot point coincides with point D , therefore the locus of point E΄ is the perpendicular E΄E on OC . For AA1 > A1D 1 is easy to see that D΄1 is on the perpendicular D΄1 E on OC .

c ) .. ( F5 , F6 ) Since the Parallel from point A 1 to OA passes through the middle of OD 1 , and in case where AOB = AOC = 90 ▫ through the middle of AD then the circle with diameter D1A passes through point D which is the base of the perpendicular , i.e.The geometrical locus of points C , E is the perpendicular CD , EE΄ .

d ) .. ( F5 , F6 )

Since A1A = A1D 1 and angle < A1AO = 60 ▫ then triangle AOD 1 is right-angle triangle and angle < D 1OA = 90 ▫ .

Since angle < AD 1O is always equal to 30 ▫ and angle D1OA is always equal to 90▫ so angle AOB is created by the rotation of right - angled triangle AOD1 through vertex O . Since tangent through Ao to circle ( O , OA΄) lies on the circle of half radius OA then this is perpendicular to OA and equal to A΄A .

F2F1 = OA A1E = OAo ─ GA1 = OA E2E1 = OA

e) .. ( F5 , F6 , F.7a )

Let point G be sliding on OB between points E and F where lines AE , AG , AF intersect circle ( O , OA ) at the points E1 , G 1 , F1 respectively and then exists FF1 > OA , GG1 = OA , E E1 < OA . Points E , F are limiting points of rotation of lines AE , AF ( because for angle < AOB = 90 ▫ → A1C = OA , A1Ao = A1E and for angle < AOB = 0 ▫ → OF = 2 . OA ) and also E1E2 = OA , F1F2 = OA and point G1 common to circle ( O , OA ) and line AG such that GG1 = OA . When point G1 is moving ( rotated ) on circle ( E2 , E2E1 = OA ) then GG1 must be stretched to G ( where then G is not on line OB ) so that GG1 = OA .

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When point G1 is moving ( rotated ) on circle ( F2 , F2F1 = OA ) then GG1 must be stretched to G ( where then G is not on line OB ) so that GG1 = OA . For both motions there is one position where point G is on line OB and is not needed GG1 to be stretched .

This position is the common point P of the two circles which is their point of intersection . At this point P exists only rotation and is not needed GG1 to be stretched . This means that point P lies on the circle ( G , GG1 = OA ) , or GP = OA .

Point A of angle < BOA is verged through two different and opposite motions , i.e.

1 . From point A΄ to point Ao where is done a parallel translation of CA΄ to the new position EAo , this is for all angles equal to 90 ▫ , and from this position to the new position EA by rotating EAo to the new position EA having always the distance E1 E2 = OA . This motion is taking place on a circle of centre E1 and radius E1 E2 . 2. From point F , where OF = 2. OA , is done a parallel translation of A΄F΄ to FAo, and from this position to the new position FA by rotating FAo to FA having always the distance F1 F2 = OA . The two motions coexist on a point P which is the point of intersection of the circles ( E2 , E2E1 = OA ) and ( F2 , F2F1 = OA ) . f) .. ( F5 , F6 , F.7a )

Remarks – Conclusions .

1 . Point E1 is common to line AE and the circle ( O,OA ) and point E2 is on line AE such that E1 E2 = OA and exists E E1 < E2 E1 Point F1 is common to line AF and the circle ( O,OA ) and point F2 is on line AF such that F1 F2 = OA and exists F F1 > F2 F1 2 . Segment E1 E2 = OA is on line EE1 and passes through A . Segment F1 F2 = OA is on line FF1 and passes through A . Segment G1 G = OA is on line GG1 and passes through A .

3 . Point G is between E , F and point G1 is common to line GA and circle ( O,OA ) When point G is on E then GG1 < E1E2 = OA and line GG1 passes through A . and when point G is on F then GG1 > F1F2 = OA and line GG1 passes through A . When point G is at such point so that GG1 = OA then point G must be at a common to the three lines EE1 , FF1 , GG1 . This is possible at the common point P of intersection of circle ( E2 , E1E2 = OA ) and ( F2 , F1F2 = OA ) , and since GG1 must be equal to OA without be stretched , then GP = OA .

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4 . On the contrary :

When circle ( G , GG1 = OA ) intersects circle ( E2 , E2 E1 = OA) then this common point is also common to the circle ( F2 , F2 F1 = OA ) . This means that the common point P of the two circles ( E2 , E2 E1 = OA) , ( F2 , F2 F1 = OA ) is constant to this motion . Since also at point P , GG1 is not needed to be stretched then GG1 = GP . Since point P is constant to this twin motion AE , AF and since only one position exists on line OB such that PG = OA then the circle ( P , PG = OA ) fixes point G such that GG1 = OA on line GA .

5. The steps of Trisection of any angle < AOB ( F6 , F7 , F8 ) 1. Draw circle ( O , OA ) and line AD parallel to OB . 2. Draw OAo ┴ OB where point Ao is on the circle ( O , OA ) and the circle ( Ao , AoE = 2.OA ) which intersects line OB at the point E . 3. Fix point F on line OB such that OF = 2 . OA 4. Draw lines AF , AE intersecting circle ( O , OA ) at points F1 , E1 respectively . 5. On lines F1F , E1E fix points F2 , E2 such that F2F1 = OA and E2 E1 = OA 6. Draw circles ( F2 , F2 F1 = OA ) , ( E2 , E2 E1 = OA) and fix point P as the common point of intersection . 7. Draw circle ( P , PG = OA ) intersecting line OB at point G and draw line GA intersecting circle ( O , OA ) at point G1 . Segment GG1 = OA . Proof : 1 . Since point P is common to circles ( F2 , F2 F1 = OA ) , ( E2 , E2 E1 = OA) , line AG between AE , AF intersects circle ( O , OA ) at the point G1 such that GG1 = OA . This common point P is such that PG = GG1 = OA . (e) 2 . Since point G1 is on the circle ( O , OA ) and since GG1 = OA then triangle GG1O is isosceles and angle < AGO = G1OG . 3 . The external angle of triangle GG1O is < AG1O = AGO + G1OG = 2. AGO . 4 . The external angle of triangle GOA is < AOB = AGO + OAG = 3. AGO .

Therefore angle < AGB = ( 1 / 3 ) . ( AOB ) ………….. ( ο.ε.δ.)

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Conclutions :

1. Following the dialectic logic of ancient Greeks ( Αναξίμανδρος ) « τό μή Ον , Ον γίγνεσθαι » ‘The Non-existent , Exist when is done ’ , ‘ The Non-existent becomes and never is ’ and the Structure of Euclidean geometry [6] in a Compact Logic Space Layer , as this exists in a known Unit ( case of 90 ▫ angle ) , then we may find a new machine that produces the 1/3 of angles .

In Euclidean geometry points do not exist , but their position and correlation is doing geometry . The universe cannot be created , because becomes and never is . According to Euclidean geometry , and since the position of points ( empty Space ) creates geometry , therefore the trisection of any angle exists in this way .

2. It has been proved [5] that two equal and perpendicular one-dimentional Units OA , OB formulate a machine which produces squares and one of them is equal tο the area of the circle ( O , OA = OB )

3. It has been proved [6] that three points formulate a Plane and from the one point passes only one Parallel to the other straight line ( three points only ) .

4. Now is proved [8] that one-dimentional Unit OA lying on two parallel lines OB , AD formulate all angles < AOB = 90 ▫ → 0 and a new geometrical machine exists which divides angle < AOB to three equal angles .

References :

[1] EUCLID’S ELEMENTS IN GREEK [2] The great text of J . L .Heiberg ( 1883-1886 ) and the English translation by Richard Fizpatrik . [3] ELEMENTS BOOK 1 . [4] GREEK MATHEMATICS by, Sir Thomas L .Heath - Dover Publications, Inc ,New York.63-3571.

[ 5] A SIMPLIFIED APPROACH OF SQUARING THE CIRCLE . ( MELAN.doc )

[ 6] THE PARALLEL POSTULATE IS DEPENDED ON THE OTHER AXIOMS ( EUCLID.doc )

[ 7] THE MEASURING OF THE REGULAR POLYGONS IN THE CIRCLE ( REGULAR.doc )

[ 8] THE TRISECTION OF ANY ANGLE . ( TRIXO.doc ) by

Markos Georgallides .

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